Chapter 2, Problem 81QRT

Interpretation Introduction

Interpretation:

The table given in the question statement has to be filled for $\text{1}\text{mol}\text{glucose,}{\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}$.

Explanation of Solution

Given the number of moles of glucose is $\text{1}$.

Number of moles of carbon $\text{=}6$ moles.

Number of moles of hydrogen $\text{=}\text{12}$ moles.

Number of moles of oxygen $\text{=}6$ mole.

Number of molecules $\text{=}$ $\text{Number}\text{of}\text{moles}\times \text{6}\text{.022}\text{×}{\text{10}}^{\text{23}}$.

Mass of carbon $\text{=}\text{12}\text{.0107}\text{g/mol}$.

Mass of hydrogen $\text{=}\text{1}\text{.0079}\text{g/mol}$.

Mass of oxygen $\text{=}15.9994\text{g/mol}$.

Molar mass of glucose $\text{=}(6\times \text{12}\text{.0107)}\text{+}\text{(}12\text{×}\text{1}\text{.0079)}\text{+}(6\times \text{15}\text{.9994)}\text{=}180.1554\text{g/mol}$.

Then the table will be:

$\begin{array}{l}{\text{C}}_6{\text{H}}_{12}{\text{O}}_6\text{Carbon}\text{Hydrogen}\text{Oxygen}\\ \text{Amount}\text{of}\text{substance}\text{(mol)}1\text{mol}6\text{mol}12\text{mol}6\text{mol}\\ \text{Number}\text{of}\text{molecules}\text{or}\text{atoms}\text{6}{\text{.022×10}}^{\text{23}}3.613{\text{×10}}^{\text{24}}7.226{\text{×10}}^{\text{24}}3.613{\text{×10}}^{\text{24}}\\ \text{Molar}\text{mass}\text{180}\text{.1554}\text{g/mol}72.0642\text{g/mol}12.0948\text{g/mol}95.9964\text{g/mol}\end{array}$